Rate of reduction of the volume of a raindrop

I have trouble solving the following question, would someone offer some help? thanks.

In very dry regions, the phenomenon called Virga is very important because it can endanger aeroplanes. [See Wikipedia]

Virga is rain in air that is so dry that the raindrops evaporate before they can reach the ground. Suppose that the volume of a raindrop is proportional to the 3/2 power of its surface area. [Why is this reasonable? Note: raindrops are not spherical, but let's assume that they always have the same shape, no mater what their size may be.]

Suppose that the rate of reduction of the volume of a raindrop is proportional to its surface area. [Why is this reasonable?]

Find a formula for the amoung of time it takes for a virga raindrop to evaporate completely, expressed in terms of the constants you introduced and the initial surface area of a raindrop. Check that the units of your formula are correct. Suppose somebody suggests that the rate of reduction of the volume of a raindrop is prpoportional to the square of the surface area. Argue that this cannot be correct.

The volume of a sphere (and a falling rain drop is very well modeled by a sphere, not a "raindrop" shape) is given by [itex]V= (4/3)\pi r^3[/itex]. Its surface are is given by [itex]A= 4\pi r^2[/itex]. If you solve the second equation for r, you get [itex]r= \sqrt{A/4\pi}[/itex] (r is positive, of course). Putting that into the first equation,
[tex]V= \frac{4}{3}\pi \left(\sqrt{\frac{A}{4\pi}}\right)^3[/tex]
[tex]= \frac{4}{3}\pi \frac{A^{3/2}}{4^{3/2}\pi^{3/2}}= \frac{1}{(3)4^{1/2}\pi^{3/2}}A^{3/2}[/tex]
and you can diferentiate that with respect to A.

Another way to do this, perhaps simpler, is to use the chain rule:
[tex]\frac{dV}{dA}= \frac{dV}{dr}\frac{dr}{dA}= \frac{\frac{dV}{dr}}{\frac{dA}{dr}}[/tex]